1 files changed, 47 insertions, 4 deletions
diff --git a/3411/CH5/EX5.12.u1/Ex5_12_u1.sce b/3411/CH5/EX5.12.u1/Ex5_12_u1.sce
index 35bb1c6fd..a07b9e963 100644
--- a/3411/CH5/EX5.12.u1/Ex5_12_u1.sce
+++ b/3411/CH5/EX5.12.u1/Ex5_12_u1.sce
@@ -2,13 +2,56 @@
 clc();
 clear;
 //To determine the cubic structure of element and lattice constant and to identify element
-ratio=(sin(20*%pi/180)/sin(29*%pi/180))^2
+//Diffraction data
+theta21=40        //units in degrees
+theta22=58        //units in degrees
+theta23=73        //units in degrees
+theta24=86.8        //units in degrees
+theta25=100.4        //units in degrees
+theta26=114.7        //units in degrees
+theta1=theta21/2      //units in degrees
+theta2=theta22/2      //units in degrees
+theta3=theta23/2      //units in degrees
+theta4=theta24/2      //units in degrees
+theta5=theta25/2      //units in degrees
+theta6=theta26/2      //units in degrees
+//sin^2(theta) values
+sin1=(sin(theta1*%pi/180))^2
+sin2=(sin(theta2*%pi/180))^2
+sin3=(sin(theta3*%pi/180))^2
+sin4=(sin(theta4*%pi/180))^2
+sin5=(sin(theta5*%pi/180))^2
+sin6=(sin(theta6*%pi/180))^2
+//sin^2(theta)/0.111 value
+temp1=sin1/sin1
+temp2=sin2/sin1
+temp3=sin3/sin1
+temp4=sin4/sin1
+temp5=sin5/sin1
+temp6=sin6/sin1
+//(h,k,l) values are determined such that the sum h^2+k^2+l^2=temp value in that manner hence we have to select the (h,k,l) values
+//(h,k,l) values
+hkl1=100    //As h^2+k^2+l^2=1
+hkl2=110    //As h^2+k^2+l^2=2
+hkl3=111    //As h^2+k^2+l^2=3
+hkl4=200     //As h^2+k^2+l^2=4
+hkl5=210       //As h^2+k^2+l^2=5
+hkl6=211       //As h^2+k^2+l^2=6
+printf("unit cell Dimensions when 2*theta=%.1f is (%d) where sin^2(theta)/0.0111 is %d\n",theta21,hkl1,temp1 )
+printf("unit cell Dimensions when 2*theta=%.1f is (%d) where sin^2(theta)/0.0111 is %d\n",theta22,hkl2,temp2 )
+printf("unit cell Dimensions when 2*theta=%.1f is (%d) where sin^2(theta)/0.0111 is %d\n",theta23,hkl3,temp3 )
+printf("unit cell Dimensions when 2*theta=%.1f is (%d) where sin^2(theta)/0.0111 is %d\n",theta24,hkl4,temp4 )
+printf("unit cell Dimensions when 2*theta=%.1f is (%d) where sin^2(theta)/0.0111 is%d\n",theta25,hkl5,temp5 )
+printf("unit cell Dimensions when 2*theta=%.1f is (%d) where sin^2(theta)/0.0111 is %d\n",theta26,hkl6,temp6 )
+
+ratio=sin1/sin2
 printf("The ratio of sin(theta)^2 values for first and second angles is %.2f\n Hence the crystal structure is bcc\n",ratio)
 lamda=0.154   //units in nm
-h=1
-k=1
+//As we have used ratio of angles of  2*theta=40 degrees and 58 degrees above we use h=1,k=1,l=0 and a^2=(lamda/2)*sqrt(sqrt(h^2+k^2+l^2)/sin^2(theta))
+h=2
+k=0
 l=0
 theta=20       //units in degrees
 a=(lamda/2)*(sqrt(sqrt(h^2+k^2+l^2)/sin(theta*(%pi/180))^2))           //units in nm
 printf("Lattice constant a=%.3fnm \n And the element is tungsten Since Tungsten has lattice constant of %.3fnm and crystallizes in bcc structure",a,a)
-//Given in textbook lattice constant as a=0.318nm but the correct answer is a=0.268nm
+//Given in textbook to find  lattice constant h=1,k=1,l=1 but the correct answer is h=2,k=0,l=0